SURFACE GROUPS IN THE GROUP OF GERMS OF ANALYTIC DIFFEOMORPHISMS IN ONE VARIABLE

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ANALYTIC DIFFEOMORPHISMS IN ONE VARIABLE

Serge Cantat, Dominique Cerveau, Vincent Guirardel, Juan Souto

To cite this version:

Serge Cantat, Dominique Cerveau, Vincent Guirardel, Juan Souto. SURFACE GROUPS IN THE GROUP OF GERMS OF ANALYTIC DIFFEOMORPHISMS IN ONE VARIABLE. L’Enseignement Mathématique , Zürich International Mathematical Society Publishing House, 2020, 66 (1-2), pp.93- 134. �10.4171/LEM/66-1/2-6�. �hal-02271428�

DIFFEOMORPHISMS IN ONE VARIABLE.

SERGE CANTAT, DOMINIQUE CERVEAU, VINCENT GUIRARDEL, AND JUAN SOUTO

ABSTRACT. We construct embeddings of surface groups into the group of germs of analytic diffeomorphisms in one variable.

1. INTRODUCTION

1.1. The main result. Let C be the field of complex numbers and Diff(C,0) the group of germs of analytic diffeomorphisms at the origin 0∈C. Choosing a local coordinateznear the origin, every element f ∈Diff(C,0)is determined by a unique power series

f(z) =a₁z+a₂z²+a₃z³+. . .+a_nzⁿ+. . . with f⁰(0) =a₁6=0 and with a positive radius of convergence

rad(f) =

lim sup

n→+∞

|a_n|^1/n −1

. (1.1)

We denote byDiff(R,0)⊂Diff(C,0)the subgroup of real germs in this chart, i.e.

witha_i∈Rfor alli∈N(this inclusion depends on the choice of the coordinate z). The main goal of this note is the following result, that answers a question raised by E. Ghys (see [8], §3.3, or also [5], Problem 4.15).

Theorem A. LetΓbe the fundamental group of a closed orientable surface, or of a closed non-orientable surface of genus ≥4. Then Γembeds in the group Diff(R,0)and in particular inDiff(C,0).

We shall present three proofs of Theorem A. For simplicity, in this introduc- tion, we restrict to the case whereΓis is the fundamental group of an orientable surface of genus 2, and we consider the presentation

Γ₂=ha₁,b₁,a₂,b₂|[a₁,b₁] = [a₂,b₂]i. (1.2)

1

Our proofs of theorem A are inspired by [3], where it is proved that a compact topological group or a connected Lie group which contains a dense free group of rank 2 contains a dense subgroup isomorphic toΓ₂.

Recently, and independently, A. Brudnyi proved a similar result for embed- dings into the group of formal germs of diffeomorphisms (see [6])

1.2. Compact groups. Let us describe the argument used in [3] to prove the following result.

Theorem 1.1 ([3]). If a compact group G contains a free group F of rank 2, then there is an embeddingρ: Γ₂→G such that F⊂ρ(Γ₂).

Proof. Denote by F_m the free group on m generators. The first ingredient is a result by Baumslag [1] saying that Γ₂ is fully residually free; this means that there exists a sequence of morphisms p_N :Γ2→F₂ which is asymptotically injective: for everyg∈Γ2\ {1}, p_N(g)6=1 ifNis large enough.

To be more explicit, we use the presentation (1.2) ofΓ₂, and we note that the subgroupha₁,b₁iofΓ₂is a free groupF₂=ha₁,b₁i. Let p:Γ₂→ha₁,b₁ibe the morphism fixinga₁andb₁and sendinga₂andb₂toa₁andb₁ respectively. Let τ:Γ₂→Γ₂be the Dehn twist around the curvec= [a₁,b₁], i.e. the automorphism that fixesa₁andb₁and senda₂andb₂toca₂c⁻¹andcb₂c⁻¹respectively.

Proposition 1.2(see [3, Corollary 2.2]). Given any g∈Γ₂\ {1}, there exists a positive integer n₀such that p◦τ^N(g)6=1for all N≥n₀.

Now, fix an embeddingι: ha₁,b₁i →Gsuch thatι(ha1,b₁i) =F. Composing p◦τ^N withι, we get a sequence of points p_N:=ι◦p◦τ^N in Hom(Γ2,G). Now, consider the element h=ι(p(c)) of G, and let T be the closure of the cyclic grouphhiin the compact groupG. Fort∈T, define a morphismρt: Γ2→Gby ρt(a1) =ι◦p(a1), ρt(a2) =t◦ι◦p(a1)◦t⁻¹, (1.3) ρ_t(b₁) =ι◦p(b₁), ρ_t(b₂) =t◦ι◦p(b₁)◦t⁻¹; (1.4) these representations are well defined and satisfy ρ_t =ι◦p◦τ^N whent =h^N. Moreover, on the subgroupha₁,b₁i,ρ_tcoincides withι◦p, soF⊂ρ_t(Γ2). Thus, (ρt)t∈T is a compact subset

R

^(T)⊂Hom(Γ2,G)that contains the sequence of pointsp_N. For everyginΓ₂\ {1}, the subset

R

^(T)g={ρ_t|ρ_t(g)6=1}is open, and Proposition 1.2 shows that it is dense because{hⁿ|n≥n₀}is dense inT for every integern₀. By the Baire theorem, the subset of injective representations ρ_t is a denseG_δin

R

^(T), and this proves Theorem 1.1.

The groupDiff(R,0)contains non-abelian free groups (this is well known, see Section 3.3), and one may want to copy the above argument forG=Diff(R,0) instead of a compact group. The Koenigs linearization theorem says that if f ∈ Diff(R,0)satisfies f⁰(0)>1, then f is conjugate to the homothety z7→ f⁰(0)z;

in particular, there is a flow of diffeomorphisms(ϕ^t)t∈R for which ϕ¹= f. In our argument, the compact group T introduced to prove Theorem 1.1 will be replaced by such a flow, hence by a group isomorphic to(R,+). Also, in that proof,h=ι(p(c))was a commutator, and the derivative of any commutator in Diff(R,0)is equal to 1 at the origin, so that Koenigs theorem can not be applied to a commutator. Thus, we need to change p_N into a different sequence of morphisms: the Dehn twistτwill be replaced by another automorphism of Γ2, twisting along three non-separating curves.

This argument will be described in details in Sections 2 and 3; the reader who wants the simplest proof of Theorem A in the case of orientable surfaces only needs to read these sections. Non orientable surfaces are dealt with in Section 4.

1.3. Lie groups. Now, let us look at representations in a linear algebraic sub- groupGofGL_m(R). Assuming that there is a faithful representationι: F₂→G with dense image, we shall construct a faithful representationΓ₂→G.

The representation variety Hom(Γ2,G)is an algebraic subset ofG⁴. Let

R

^be

the irreducible component containing the trivial representation. Let p_N: Γ2→ F₂ be an asymptotically injective sequence of morphisms, as given by Baum- slag’s proposition. When the image ofρis dense, one can prove thatι◦p_N is in

R

for arbitrarily large values ofN. Forg∈Γ₂\ {1}, the subset

R

g⊂

R

^{of ho-}

momorphisms killinggis algebraic, and it is a proper subset because it does not containι◦p_N for some largeN. Then, a Baire category argument in

R

^implies

that a generic choice ofρ∈

R

is faithful.

To apply this argument to G= Diff(R,0), one needs a good topology on Diff(R,0), and a good “irreducible variety”

R

^⊂Hom(Γ2,G)containingι◦p_N, in which a Baire category argument can be used. This approach may seem diffi- cult because Hom(Γ2,G)is a priori far from being an irreducible analytic variety but, again, the Koenigs linearization theorem will provide the key ingredient.

First, we shall adapt an idea introduced by Leslie in [15] to define a useful group topology onDiff(R,0)(see Section 5). With this topology,Diff(R,0)is an increasing union of Baire spaces, which will be enough for our purpose. Denote byCont(R,0)⊂Diff(R,0)the set of elements f ∈Diff(R,0)with|f⁰(0)|<1;

Contstands for “contractions”. Consider the set

R

of representationsρ: Γ₂→ Diff(R,0)withρ(a₁)tangent to the identity, andρ(b₁)∈Cont(R,0). Then, the key fact is that the map

Ψ:

R

^{→Cont(R,0)×Diff(R,0)×Diff(R,0)}

ρ7→(ρ(b₁),ρ(a₂),ρ(b₂))

is a continuous bijection. Indeed, the defining relation of Γ is equivalent to a₁b₁a⁻¹₁ = [a₂,b₂]b₁. Given (g₁,f₂,g₂)∈Cont(R,0)×Diff(R,0)×Diff(R,0), the germsg₁ and[f₂,g₂]g₁have the same derivative at the origin and, from the Koenigs linearization theorem, there is a unique f₁∈Diff(R,0) tangent to the identity solving the equation f₁g₁f₁⁻¹= [f₂,g₂]g1: by construction there is a unique morphism ρ: Γ₂→Diff(R,0) that maps the a_i to the f_i, and the b_i to theg_i, and this representation satisfiesΨ(ρ) = (g₁,f₂,g₂). With this bijectionΨ and the topology of Leslie, we can identify

R

with a union of Baire spaces, in which the Baire category argument applies.

1.4. Other fields. Letkbe a finite field withpelements. The groupDiff¹(k,0), also known as the Nottingham group, is the group of power series tangent to the identity and with coefficients in the finite field k. It is a compact group containing a free group (see [22]). Thus, by [3], it contains a surface group.

Now, let pbe a prime number, and letQ_pbe the field of p-adic numbers.

Consider the subgroupDiff¹(Zp,0)⊂Diff(Qp,0)of formal power series tan- gent to the identity and with coefficients inZ_p. First, note that all elements f ofDiff¹(Zp,0)satisfy rad(f)≥1, so thatDiff¹(Zp,0)acts faithfully as a group of (p-adic analytic) homeomorphisms on {z∈Z_p ; |z|< 1}. So, in that re- spect, Diff¹(Z_p,0)is much better than the group of germs of diffeomorphisms Diff(C,0). Moreover, with the topology given by the product topology on the coefficientsa_n∈Z_pof the power series, the groupDiff¹(Zp,0)becomes a com- pact group. And this compact group contains a free group. By the result of [3]

described in Section 1.2, it contains a copy of the surface groupΓ₂. So, we get a surface group acting faithfully as a group ofp-adic analytic homeomorphisms on{z∈Z_p; |z|<1}. In Section 7 we give a third proof of Theorem A that starts with the case of p-adic coefficients.

1.5. Organisation. The article is split in four parts.

I.– Sections 2 to 4 give a first proof of Theorem A; Section 4 is the only place where we deal with non-orientable surfaces. We refer to Theo- rem B in Section 4.3 for a stronger result, in which the fieldRis replaced by any non-discrete, complete valued fieldk.

II.– Section 5 and 6 present our second proof, based on the construction of a group topology onDiff(C,0).

III.– Then, our p-adic proof is presented in Section 7.

IV.– Section 8 draws some consequences and list a few open problems, while the appendix shows how to construct free groups inDiff(C,0), orDiff(k,0) for any non-discrete and complete valued field.

Acknowledgement. Thanks to Yulij Ilyashenko, Frank Loray and Ludovic Mar- quis for several discussions on this subject, and to the participants of the sem- inars of Dijon, Moscou, Paris, and Toulouse during which the results of this paper where presented.

CONTENTS

1. Introduction 1

2. Germs of diffeomorphisms and the Koenigs Linearization Theorem 5

3. Embedding orientable surface groups 7

4. Non-orientable surface groups 13

5. The final topology on germs of diffeomorphisms 18 6. A large irreducible component of the representation variety 27

7. A p-adic proof 31

8. Complements and open questions 34

9. Appendix: Free groups 37

References 40

– Part I. –

2. GERMS OF DIFFEOMORPHISMS AND THEKOENIGSLINEARIZATION

THEOREM

2.1. Formal diffeomorphisms. Let k be a field (of arbitrary characteristic).

Denote bykJzKthe ring of formal power series in one variable with coefficients

ink. For every integern≥0, letA_n: kJzK→kdenote the n-th coefficient func- tion:

A_n: f =

∑

^{anzn 7→ An(f) =an. (2.1)}

Aformal diffeomorphismis a formal power series f ∈kJzKsuch thatA₀(f) =0 andA₁(f)6=0. The composition f◦gdetermines a group law on the set

Diff(k,d 0) ={f ∈kJzK|A₀(f) =0 andA₁(f)6=0} (2.2) of all formal diffeomorphisms.

For eachn≥1, there is a polynomialP_n∈Z[A₁,B₁, . . . ,A_n,B_n]such that if f =

∑a_nzⁿandg=∑b_nzⁿthen f◦g=∑_n≥1P_n(a₁,b₁, . . . ,a_n,b_n)zⁿ. Similarly, there are polynomialsQ_n∈Z[A1, . . . ,A_n][A⁻¹₁ ]such that f⁻¹=∑_n≥1Q_n(a1, . . . ,a_n)zⁿ if f =∑a_nzⁿ; the polynomial function Q_n is given by the following inversion formula:

1 aⁿ₁

∑

k1,k2,...

(−1)^k1+k2+...·(n+1)· · ·(n−1+k₁+k₂+. . .) k₁!k₂!· · · ·

a₂ a₁

k₁ a₃ a₁

k₂

· · ·

wherea_i=A_i(f)and the sum is over all sequences of integersk_isuch that k₁+2k₂+3k₃+· · ·=n−1.

We refer to [13] where this is proved for f and g tangent to the identity; the general case easily follows.

To encapsulate this kind of properties, we introduce the following definition.

Let mbe a positive integer. By definition, a function Q: Diff(k,d 0)^m→k is a polynomial function with integer coefficients, if there is an integer n, and a polynomialq∈Z[A1,1,A_2,1, . . . ,A_m−1,n,A_m,n][A⁻¹_1,1, . . . ,A⁻¹_m,1]such that

Q(f₁, . . . ,f_m) =q(A1(f₁), . . . ,A_n(f_m)) (2.3) for allm-tuples(f₁, . . . ,f_m)∈Diff(k,d 0)^m; we denote byZ[Diff(k,d 0)^m]this ring of polynomial functions.

Let F_m = he₁, . . . ,e_mi be the free group of rank m. To every word w =

eⁿ_i¹

1 . . .eⁿ_i^k

k inF_m, we associate theword mapw:Diff(k,d 0)^m→Diff(k,d 0), (g₁, . . . ,g_m)7→w(g₁, . . . ,g_m)^def= gⁿ_i¹

1 ◦. . .◦gⁿ_i^k

k. (2.4)

Since composition and inversion are polynomial functions onDiff(k,d 0), we get:

Lemma 2.1. Let w:Diff(k,d 0)^m→Diff(k,d 0) be the word map given by some element of the free groupF_m. For each n≥1, there is a polynomial function Q_w,n∈Z[Diff(k,d 0)^m]such that

A_n(w(g₁, . . . ,g_m)) =Q_w,n(g₁, . . . ,g_m) for all g₁, . . . ,g_m∈Diff(k,d 0).

2.2. Diffeomorphisms and Koenigs linearization Theorem. Suppose now that k is endowed with an absolute value| · |: k→R₊. Then kbecomes a metric space with the distance induced by| · |. We shall almost always assume that

• kis not discrete, equivalently there is an elementx∈kwith|x| 6=0,1;

• kis complete.

Letk{z}be the subring ofkJzKconsisting of power series f(z) =∑a_nzⁿwhose radius of convergence rad(f)is positive (see Equation (1.1)). Whenkis com- plete, the series∑a_nzⁿconverges uniformly in the closed diskDr={z∈k| |z| ≤ r}for everyr<rad(f). The group of germs of analytic diffeomorphisms is the intersectionDiff(k,0):=Diff(k,d 0)∩k{z}; it is a subgroup ofDiff(k,d 0).

A germ f ∈Diff(k,0) is hyperbolic if|A₁(f)| 6=1. The following result is proved in [18, Chapter 8] and [11, Theorem 1, p. 423] (see also [20, Theorem 1] or [14]).

Theorem 2.2(Koenigs linearization theorem). Let(k,| · |)be a complete, non- discrete valued field. Let f ∈Diff(k,0)be a hyperbolic germ of diffeomorphism.

There is a unique germ of diffeomorphism h∈Diff(k,0) such that h(f(z)) = A₁(f)·h(z)and A₁(h) =1.

3. EMBEDDING ORIENTABLE SURFACE GROUPS

3.1. Abstract setting. Our strategy to construct embeddings of surface groups relies on the following simple remark. Let Γ be a countable group, and G be any group. Consider a non-empty topological space

R

, with a mapΦ:s∈

R

^7→

Φ_s∈Hom(Γ,G). Giveng∈Γ, set

R

g={s∈

R

^|Φs(g) =1}.

Lemma 3.1. Assume that

R

has the following 3 properties:

(1) Baire:

R

is a Baire space;

(2) Separation:for every g6=1inΓ,Φ_s(g)6=1for some s∈

R

^;

(3) Irreducibility:for every g∈Γ, either

R

g=

R

^or

R

gis closed with empty interior.

Then the set of s∈

R

^{such thatΦ}sis an injective homomorphism is a dense G_δ in

R

; in particular, it is non-empty.

Proof. For anyg∈Γ\ {1}, one has

R

g6=

R

^{by (2), so}

R

gis closed with empty interior by (3). By the Baire property,

R

^\(∪g∈Γ\{1}

R

g) is a dense G_δ. But

R

^\(∪g∈Γ\{1}

R

g)is precisely the set ofs∈

R

^{such thatΦ}s is injective.

3.2. Baumslag Lemma. As explained in the introduction, it is proved in [1]

that the fundamental group of an orientable surface is fully-residually free. We need a precise version of this result; to obtain it, the main technical input is the Baumslag’s Lemma (see [19, Lemma 2.4]):

Lemma 3.2(Baumslag’s Lemma). Let n≥1be a positive integer. Let g₀, . . ., g_nbe elements ofF_k, and let c₁,. . ., c_nbe elements ofF_k\ {1}. Assume that for all1≤i≤n−1, g⁻¹_i c_ig_idoes not commute with c_i+1. Then for N large enough,

g₀c^N₁g₁c^N₂ . . .c^N_n−1g_n−1c^N_ng_n6=1.

Sketch of proof (I). The groupPSL₂(R)acts on the hyperbolic planeHby isome- tries, and contains a subgroupΓsuch that (0) Γis isomorphic toF_k, (1) every elementg6=Id inΓis a loxodromic isometry ofH, and (2) two elementsgand hinΓ\ {Id}commute if and only if they have the same axis, if and only if they share a common fixed point on∂H. One can find such a group in any lattice of PSL₂(R). To prove the lemma, we prove it inΓ.

Fix a base pointx∈H, denote byα_iandω_ithe repulsive and attracting fixed points ofc_iin∂H, and consider the word

g₀c^N₁g₁c^N₂g₂.

Formlarge enough, c^m₂g₂ mapsx to a point which is near ω₂. If g₁(ω2) were equal toα₁, thenc₁ andg₁c₂g⁻¹₁ would share the common fixed pointα₁, and they would commute. Thus, g₁(ω2)6=α₁ and then g₀c^m₁⁰g₁c^m₂g₂ maps x to a point which is nearg₀(ω1)ifm⁰is large enough. Thus, g₀c^N₁g₁c^N₂g₂(x)6=xfor largeN. The proof is similar ifnis larger than 2.

Sketch of proof (II). We rephrase this proof, using the action ofF_kon its bound- ary, because this boundary will also be used in the proof of Proposition 3.3.

Fix a basisa₁, . . . ,a_k ofF_k, and denote by∂F_k the boundary ofF_k. The ele- ments of∂F_k are represented by infinite reduced words in the generatorsa_i and their inverses. Ifgis an element ofF_kandαis an element of∂F_kthe concatena- tiong·αis an element of∂F_k: this defines an action ofF_k by homeomorphisms

on the Cantor set∂F_k. Ifgis a non-trivial, its action on∂F_khas exactly two fixed points, given by the infinite wordsω(g) =g· · ·g· · · andα(g) =g⁻¹· · ·g⁻¹· · · (there are no simplifications if gis given by a reduced and cyclically reduced word). Then we get: (1) every elementg6=Id inF_k has a north-south dynamics on∂F_k, every orbitgⁿ·βconverging toω(g), except whenβ=α(g), and (2) two elementsgandhinF_k\ {Id}commute if and only if they have the same fixed points, if and only if they share a common fixed point on ∂F_k. One can then repeat the previous proof with the action ofF_kon its boundary.

α0

η2

α1 α2

η1

¯ η1

¯ η2

a0=a0=α0

a1=η1α1η₁⁻¹

¯

a2= ¯η2α2η¯₂⁻¹ t1=η1η¯₁⁻¹

¯

a1= ¯η1α1η¯₁⁻¹ a2=η2α2η₂⁻¹

t2=η2η¯₂⁻¹

FIGURE1. The fundamental groupΓ2.– Theαiare three loops, while theηj andηj are four paths. The figure is symmetric with respect to the plane cutting the surface along the loopsαi.

Write the surface of genus 2 as the union of two pairs of pants as in Figure 1, with respective fundamental groups

ha₀,a₁,a₂|a₀a₁a₂=1i and ha₀,a₁,a₂|a₀a₁a₂=1i. (3.1) This gives the presentation

Γ2=

*a₀,a₁,a₂, a₀,a₁,a₂,

t₁,t₂

a₀a₁a₂=1, a₀a₁a₂=1,

a₀=a₀, a₁=t₁⁻¹a₁t₁, a₂=t₂⁻¹a₂t₂ +

(3.2) which can be rewritten as

Γ₂=ha₀,a₁,a₂,t₁,t₂|a₀a₁a₂=1, a₀t₁⁻¹a₁t₁t₂⁻¹a₂t₂=1i. (3.3) Denote byp:Γ₂→ha₀,a₁,a₂i 'F₂the morphism defined byp(ai) =a_i, p(ai) = a_i, andp(t₁) =p(t₂) =1. Letτ:Γ₂→Γ₂be the (left) Dehn twist along the three curvesa₀, a₁, anda₂, i.e. the automorphism fixinga_i and sendingt_i toa_it_ia⁻¹₀ fori=1,2. Note the following facts:

• τsendsa_itoa₀a_ia⁻¹₀ ; in particular, ifgis a word in thea_i, thenτ^N(g) = a^N₀ga^−N₀ ;

• p◦τ^N fixesa_ifor everyi=0, 1, 2, and

p◦τ^N(t_j) =a^N_ja^−N₀ (3.4) for j=1, 2.

Proposition 3.3. For every g∈Γ₂\ {1}, there exists a positive integer n₀ such that p◦τ^N(g)6=1for all N≥n₀.

Proof. To uniformize notations, we definet₀=1 so that for alli∈ {0,1,2}the relation t_ia_it_i⁻¹=a_i holds, and τ mapst_i to a_it_ia⁻¹₀ . Let A=ha₀,a₁,a₂i and A=ha₀,a₁,a₂i. Writegas a shortest possible word of the following form:

g=g₀t_i1g₁t_i⁻¹

2 g₂t_i3. . .g_n−1t_i⁻¹

n g_n (3.5)

wherenis even,i_k∈ {0,1,2}for allk≤n,g_k∈Aforkeven,g_k∈Aforkodd, and the exponent oft_ikis(−1)^k+1(we allowg_k=1). One easily checks thatgcan be written in this form because all generators can (for instancet₁=1·t₁·1·t₀⁻¹·1).

Ifk is such thati_k =i_k+1, theng_k ∈ ha/ _iki ifk is even (resp g_k∈ ha/ _ikiif k is odd) as otherwise, one could shorten the word using the relationt_ika_ikt_i⁻¹

k =a_ik. First claim.If k∈ {2, . . . ,n−2}is even, g⁻¹_k a_ikg_k does not commute to a_ik+1. Ifi_k6=i_k+1, this is becauseg_k∈A'F₂and no pair ofA-conjugates ofa_ik and a_ik+1 commute. Ifi_k=i_k+1, theng_k∈ ha/ _ikias we have just seen; sincea_ikis not a proper power inA, this shows thatg_k.(a^+∞_i

k )6=a^+∞_i

k in the boundary at infinity of the free groupA, sog⁻¹_k a_ikg_kdoes not commute witha_ik, and the claim follows.

Similarly, using the fact thatg_k∈Afor odd indices, we obtain:

Second claim. If k≤n−1is odd, g⁻¹_k a_ikg_kdoes not commute to a_ik+1. We have τ^N(g_k) =g_k if k is even, andτ^N(g_k) =a^N₀g_ka^−N₀ ifk is odd. After simplifications, one has

τ^N(g) =g₀a^N_i

1t_i1g₁t_i⁻¹

2 a^−N_i

2 g₂a^N_i

3t_i3. . .g_n−1t_i⁻¹

n a^−N_i

n g_n. (3.6) Forkodd, denote byg⁰_k∈F_r the image ofg_kunder p. Applying p, we thus get

p◦τ^N(g) =g₀a^N_i1g⁰₁a^−N_i

2 g₂a^N_i3g⁰₃. . .g⁰_n−1a^−N_i

n g_n, (3.7)

withg⁰_i:=p(g_i). Let us check that the hypotheses of the Baumslag Lemma 3.2 apply. For k even, the first claim shows that g⁻¹_k a_ikg_k does not commute to

a_ik+1, as required. For k odd, we use that p is injective on A and that A con- tainsg⁻¹_k a_ikg_kanda_ik+1, and we apply the second claim to deduce thatg⁰⁻¹_k a_ikg⁰_k does not commute to a_ik+1. Applying Baumslag’s Lemma, we conclude that

p◦τ^N(g)6=1 forN large enough.

3.3. Embeddings of free groups. The group Diff(R,0) contains non-abelian free groups. This has been proved by arithmetic means [25, 10], by looking at the monodromy of generic polynomial planar vector fields [12], and by a dynamical argument [17]. We shall need the following precise version of that result.

Theorem 3.4. Let (k,| · |) be a complete non-discrete valued field. For every pair(λ1,λ2) ink^∗, there exists a pair f₁, f₂∈Diff(k,0) that generates a free group and satisfies f₁⁰(0) =λ₁and f₂⁰(0) =λ₂.

This result is proved in [2, Proposition 4.3] for generic pairs of derivatives (λ1,λ2). We provide a proof of Theorem 3.4 in the Appendix, extending the argument of [17]. We refer to Section 7.1 below for other approaches.

3.4. Embedding orientable surface groups. We can now prove Theorem A for orientable surfaces:

Theorem 3.5. LetΓ_g be the fundamental group of a closed, orientable surface of genus g. Then, there exists an injective morphismΓ_g→Diff(R,0).

The groupΓ₀ is trivial. The group Γ₁ is isomorphic toZ², hence it embeds in the group of homothetiesz7→λz, λ∈R^∗₊. If g≥2, then Γ_g embeds in Γ₂. To see this, fix a surjective morphism Γ₂→Z, and take the preimage Λ⊂Γ₂ of the subgroup (g−1)Z⊂Z. Then, Λ is a normal subgroup of indexg−1 inΓ2, and it is the fundamental group of a closed surfaceΣ, given by a Galois cover of degreeg−1 of the surface of genus 2. Since the Euler characteristic is multiplicative, the genus ofΣsatisfies−2(g−1) =2−2g(Σ). Thus, g(Σ) =g andΛis isomorphic toΓ_g. Thus, we now restrict to the caseg=2.

By Theorem 3.4, we can fix an injective morphism

ρ₀:F₂=ha₀,a₁,a₂|a₀a₁a₂=1i→Diff(R,0) (3.8) such that the images f₁=ρ₀(a₁), f₂=ρ₀(a₂), and f₀=ρ₀(a₀) =f₂⁻¹f₁⁻¹satisfy f₁⁰(0) =λ₁>1, f₂⁰(0) =λ₂>1, f₀⁰(0) =λ₀<1 (3.9)

for some real numbers λ₁ and λ₂ > 1 and λ₀ = (λ1λ₂)⁻¹. In particular, f₀, f₁, and f₂ are hyperbolic. For λ∈R^∗, denote by m_λ(z) =λz the correspond- ing homothety. For i∈ {0,1,2}, the Koenigs linearization theorem shows that f_i is conjugate to the homothety m_λi: there is a germ of diffeomorphism h_i ∈ Diff(R,0) such that f_i =h_i◦m_λi◦h⁻¹_i . Thus f_i extends to the multiplicative flowϕ_i:R^∗₊→Diff(R,0)defined byϕ^s_i =h_i◦m_s◦h⁻¹_i fors∈R^∗₊; by contruc- tion, ϕ^λ_iⁱ = f_i and ϕ^s_i commutes with f_i for all s>0. We note that s7→ϕ^s_i is polynomial in the sense that for allk∈N, s7→A_k(ϕ^s_i)is a polynomial function with real coefficients in the variablessands⁻¹.

Set

R

^{= (R∗}+)³. As in Section 3.2, consider the presentation

Γ2=ha₀,a₁,a₂,t₁,t₂|a₀a₁a₂=1, a₀t₁⁻¹a₁t₁t₂⁻¹a₂t₂=1i. (3.10) Givens= (s₀,s₁,s₂)∈(R^∗₊)³, we define a morphismΦ_s:Γ₂→Diff(R,0)by

Φs(ai) = f_i fori∈ {0,1,2} (3.11) Φ_s(t_i) =ϕ^s_iⁱϕ^s₀⁰ fori∈ {1,2} (3.12) This provides a well defined homomorphism becauseϕ_icommutes with f_i. As we shall see below, this morphismΦ_sis constructed to coincide withρ₀◦p◦τ^N fors= (λ^N₀,λ^N₁,λ^N₂)(see Equation (3.4)).

Remark 3.6. For every s∈

R

, the image of Φ_s contains f₁ and f₂, hence the free groupρ0(F2). This will be used in Section 4.3.

To conclude, we check that the three assumptions of Lemma 3.1 hold for this family of morphisms(Φs)_s∈R.

Clearly,

R

is a Baire space.

To check the irreducibility property, considerg∈Γ₂and assume that

R

g6=

R

^:

this means that there exists a parameter s∈

R

and an index k≥ 1 such that A_k(Φs(g))6=A_k(Id). The maps= (s0,s₁,s₂)7→A_k(Φs(g))−A_k(Id)is a polyno- mial function in the variabless^±1₀ ,s^±1₁ , ands^±1₂ that does not vanish identically on

R

, so its zero set is a closed subset with empty interior.

We now check that

R

has the separation property. As in Section 3.2, denote by p:Γ₂→F₂=ha₀,a₁,a₂|a₀a₁a₂=1i the morphism obtained by killingt₁

andt₂. For the parameters= (1,1,1),Φ_sis equal toρ₀◦p. More generally, set- tings_N = (λ^N₀,λ^N₁,λ^N₂)forN∈N, the morphismΦ_sN :Γ₂→Diff(R,0)satisfies Φ_sN(a_i) = f_i fori∈ {0,1,2} (3.13)

Φs_N(ti) =ϕ^u

N i

i ϕ^u

N 0

0 = f_i^Nf₀^N fori∈ {1,2}. (3.14) This means that Φ_sN =ρ0◦p◦τ^N where, as in Section 3.2, τ:Γ2→Γ₂ is the Dehn twist along the three curves a_i. By Proposition 3.3, for all g∈Γ2\ {1}

there existsN∈Nsuch thatp◦τ^N(g)6=1. Sinceρ₀is injective, this implies that Φ_sN(g)6=1 which shows that

R

has the separation property.

4. NON-ORIENTABLE SURFACE GROUPS

Theorem 4.1. Let N_gbe the fundamental group of a closed non-orientable sur- face of genus g≥4. There exists an injective morphism N_g→Diff(R,0).

Remark 4.2. The fundamental groupN₃of the non-orientable surface of genus 3 is not fully residually free, and our methods do not apply to this group. (See [16], Proposition 9.)

4.1. Even genus. We first treat the case of an even genus g≥4. In this case, the groupN_gembeds inN₄. Indeed, the non-orientable surface of genus 4 is the connected sum of a torusR²/Z² with two projective planes P²(R). Taking a cyclic cover of the torus of degreek, we get a surface homeomorphic to the con- nected sum ofR²/Z²with 2kcopies ofP²(R), hence a non-orientable surface of genus 2(k+1). Thus, it suffices to prove thatN₄embeds inDiff(R,0).

The non-orientable surface of genus 4 is homeomorphic to the connected sum of 4 copies ofP²(R), and this gives the presentation (see Figure 2)

N₄=ha₁,a₂,b₁,b₂|a²₁a²₂b²₂b²₁=1i. (4.1) Let p:N₄→ha₁,a₂ibe the morphism fixing a₁,a₂ and sending b₁ and b₂ to a⁻¹₁ anda⁻¹₂ respectively. Let τ:N₄→N₄ be the Dehn twist around the curve γ= (a²₁a²₂)⁻¹, i.e. the automorphism that fixesa₁anda₂ and sendsb₁andb₂ to γb₁γ⁻¹andγb₂γ⁻¹respectively.

Lemma 4.3. Given any g∈N₄\ {1}, there exists n₀∈Nsuch that for all N≥n₀, p◦τ^N(g)6=1.

For the proof. The proof of this statement is completely analogous to the proof of [3, Corollary 2.2], using Baumslag Lemma, we leave it as an exercise to the

reader. See also [9, Proposition 4.13].

b²₁

a1 a2

b2

b1

a²₂

γ=b²₂b²₁= (a²₁a²₂)⁻¹ a²₁

b²₂

FIGURE2. The fundamental groupN₄.– The base point is represented by •, the 4 generators are a₁, a₂, b₁, b₂, and the curve γ is used to construct the Dehn twistτ.

Using Theorem 3.4, we fix two germs of diffeomorphisms f₁and f₂∈Diff(R,0) generating a free group and satisfying f₁⁰(0)>1 and f₂⁰(0)>1. We denote by

ρ0:F₂=ha₁,a₂i→Diff(R,0) (4.2) the injective morphism sendinga_ito f_ifori∈ {1,2}. In particular,

ρ₀(γ) = (f₁²◦ f₂²)⁻¹ (4.3) is a hyperbolic germ: its derivativeλ= ((f₁²◦ f₂²)⁰(0))⁻¹ is<1. The Koenigs linearization theorem gives an elementh∈Diff(R,0)such thatρ₀(γ) =h◦m_λ◦ h⁻¹. Consider the multiplicative flow ϕ:R^∗₊→Diff(R,0) defined byϕ^s =g◦ m_s◦g⁻¹. As above, ϕ^λ = ρ0(γ), ϕ^s commutes with ρ0(γ) for all s>0, and s7→ϕ^s is a polynomial map: for all k∈N, s7→A_k(ϕ^s) is a polynomial in the variablessands⁻¹.

Set

R

^=R∗+. Given s∈R^∗₊, consider the morphism ρ_s :N₄→ Diff(R,0) defined by

a₁7→ f₁ a₂7→ f₂

b₁7→ϕ^sf₁⁻¹ϕ^−s b₂7→ϕ^sf₂⁻¹ϕ^−s.

This gives a well defined homomorphism becauseϕ^scommutes with f₁²f₂². We now check the three assumptions of Lemma 3.1. Clearly,

R

^{is a Baire}

space. The irreducibility is a consequence of the fact that for anyg∈N₄, and anyk∈Nthe maps7→A_k(ϕs(g))is a polynomial function in the variabless^±1. The separation property follows from Lemma 4.3 together with the fact that ρ_λ^N =ρ₀◦p◦τ^N and thatρ₀is injective.

4.2. Odd genus. We now treat the case of a non-orientable surface of odd genus g=2k+1,k≥2. One can writeN_2k+1as (see Figure 3 below)

N_2k+1=ha₁, . . . ,a_k,c,b₁, . . . ,b_k|a²₁. . .a²_kc²b²_k. . .b²₁=1i. (4.4) This group splits as a double amalgam of free groups

N_2k+1=ha₁, . . .a_ki ∗

a²₁...a²_k=γ⁻¹hγ,ci ∗

c⁻²γ=b²_k...b²₁

hb₁, . . . ,b_ki. (4.5) We shall use the following notation to refer to this amalgam structure:

• A₁=ha₁, . . .a_kiande_1,2= (a²₁. . .a²_k)⁻¹;

• A₂=hγ,ciande_2,1=γande_2,3=c⁻²γ=δ;

• A₃=hb₁, . . . ,b_kiande_3,2=b²_k. . .b²₁.

So, each of theA_i is a free group and the amalgamation is given bye_1,2=e_2,1 ande_2,3=e_3,2.

Define a morphism p: N_2k+1→ha₁, . . . ,a_ki 'F_k by

a_i7→a_i fori≤k c7→a⁻²_k b_i7→a⁻¹_i fori≤k−1 b_k7→a_k (the structure of almagam shows that pis well defined).

a1 a2 ak

b1 b2 bk

c γ

c² δ=c⁻²γ

b²₁

a²₁ a²_k

b²_k

FIGURE 3. The fundamental groupN_2k+1.

Lemma 4.4. The morphism p: N_2k+1→F_k is injective in restriction to each of the three subgroups of the amalgam(4.5).

Proof. By construction, it is injective in restriction toha₁, . . . ,a_kiand in restric- tion tohb₁, . . . ,b_ki. Then, note that p(hγ,ci) =ha²₁. . .a²_k,a⁻²_k iis isomorphic to F₂because it is a non-abelian subgroup of a free group. SinceF₂is Hopfian, p

is necessarily injective in retriction tohγ,ci.

Considerδ=b²_k· · ·b²₁=c⁻²γand note that p(δ) =a²_ka⁻²_k−1· · ·a⁻²₁ . Letτbe the Dehn twist corresponding to the decomposition above, i.e. the automorphism fixing a_i, sending c to γcγ⁻¹ and sending b_i to (γδ)bi(γδ)⁻¹. Since τ is the composition of the twists given byγandδand these two twists commute we get

τ^N(b) = (γ^Nδ^N)b(γ^Nδ^N)⁻¹, ∀b∈A₃.

In this situation, one can prove the following lemma in a similar way to Propo- sition 3.3.

Lemma 4.5. Given any g∈N_2g+1\ {1}, there exists n₀ ∈N such that for all N≥n₀, p◦τ^N(g)6=1.

Proof. Writegas a word in the graph of groups, i.e.g=s₀. . .s_n withs_k ∈A_rk (we allows_k=1) for some r_k ∈ {1,2,3}, withr_k+1=r_k±1, and r₀=r_n=1.

We take this word of minimal possible length among words satisfying these contraints. Ifkis such thatr_k−1=r_k+1, thens_k∈ he/ _rk,rk+1isince otherwise, one could shorten the word using the structure of amalgam (in particulars_k 6=1 in this case). Now one easily checks that

τ^N(g) =s₀d₁^ε1Ns₁d₂^ε2Ns₂· · ·d_n^εnNs_n (4.6) whered_k=e_rk−1,rk∈ {γ,δ}, andε_k=r_k−r_k−1∈ {±1}.

We claim thats⁻¹_k d_ks_kdoes not commute withd_k+1. Ifd_k6=d_k+1, this follows from the fact thatγcommutes with no conjugate ofδinA₂=hc,δi. Ifd_k=d_k+1, thenr_k−1=r_k+1, sos_k∈ he/ _rk,rk+1i=hd_ki. If[s⁻¹_k d_ks_k,d_k] =1, thens_kpreserves the axis ofd_k in the Cayley graph of the free groupA_rk, sos_k is a power ofd_k, becaused_k∈ {γ,δ}is not a proper power; this contradicts thats_k∈ hd/ _ki.

Denote by s_k, d_k ∈F_r the images of s_k, d_k under p. Since pis injective on eachA_rk, s⁻¹_k d_ks_k does not commute with s_k+1, so the hypotheses of Baumslag Lemma apply to the word

p◦τ^N(g) =d^ε₀^0Ns₁d^ε₁^1Ns₂. . .d^ε_n−1^n−1Ns_nd^ε_n^nN (4.7)

so p◦τ^N(g)6=1 forN large enough.

Now considerkelements f₁, . . ., f_k of Diff(R,0)generating a free group of rankkwith f_i⁰(0)>1 for alli∈ {1, . . . ,k}, and f_k⁰(0)< f₁⁰(0). Such a set can be obtained from two generatorsg₁andg₂of a free group of rank 2 withg⁰_i(0)>1, as in Theorem 3.4, by taking f_i=gⁱ₁◦g²₂◦g⁻ⁱ₁ fori<kand f_k=g^k₁◦g₂◦g^−k₁ . Let ρ0:F_k=ha₁, . . . ,a_ki→Diff(R,0)be the injective morphism sendinga_ito f_ifor i≤k. In particular,ρ₀(γ) = (f₁²◦ · · · ◦f_k²)⁻¹andρ₀(p(δ)) = f_k²◦f_k−1⁻² ◦ · · · ◦f₁⁻² are hyperbolic. Using Koenigs linearization theorem as above, there exists two multiplicative flowsϕandψ:R^∗₊→Diff(R,0)and a pair of positive real numbers λandµsuch that (1)ϕ^λ=ρ₀(γ)andψ^µ=ρ₀(p(δ)), and (2)s7→ϕ^sands7→ψ^s are polynomial mappings.

Set

R

^{= (R∗}+)²and, for every(s,s⁰)∈

R

, define a morphismρ_s,s⁰:N_2k+1→Diff(R,0) by

a_i7→ f_i fori≤k c7→ϕ^sf_k⁻²ϕ^−s b_i7→ϕ^sψ^s

0f_i⁻¹(ϕ^sψ^s

0)⁻¹ fori≤k−1 b_k7→ϕ^sψ^s

0f_k(ϕ^sψ^s

0)⁻¹ (it is well defined because ϕ^s and ψ^s0 commute with ρ₀(γ) = (f₁²◦ · · · ◦ f_k²)⁻¹ andρ0(p(δ)) = f_k²◦f_k−1⁻² ◦ · · · ◦f₁⁻²respectively).

The assumptions of Lemma 3.1 hold:

R

is a Baire space, and the irreducibil- ity follows from the fact that the mapss7→ϕ_sands⁰7→ϕ_s⁰are polynomials in the variables s^±1,s^0±1. The separation property follows from Lemma 4.5 together with the fact thatρ_λN,µ^N =ρ₀◦p◦τ^N, and thatρ₀is injective.

4.3. Embeddings in Diff(k,0). The proofs just given provide the following statement.

Theorem B. Let(k,| · |)be a non-discrete and complete valued field.

(1) Let Γ be the fundamental group of a closed orientable surface, or a closed non-orientable surface of genus≥4. Then, there is an embedding ofΓintoDiff(k,0).

(2) Let F ⊂Diff(k,0) be a free group of rank2, generated by two germs f and g with|f⁰(0)|>1 and|g⁰(0)|>1. Then, there is an embedding ofΓ₂, the fundamental group of a closed, orientable surface of genus2, intoDiff(k,0)whose image contains F.

Proof. For the first assertion, we just have to replaceR by k in the proofs of Theorem 3.5 and 4.1. The parameter space is

R

^{= (k∗)3ork∗ or(k∗)2, and it}

is a Baire space because(k,| · |)is complete.